本地测都能过，上来全re

#include <bits/stdc++.h>

using namespace std;

struct Tree{
	int left, right;
	int sum, delta;
};
Tree tree[4 * 200010];
int a[100010];
int n, t;

void build(int id, int l, int r)
{
	tree[id].left = l;
	tree[id].right = r;
	if (tree[id].left == tree[id].right)
	{
		tree[id].sum = a[l];
	}
	else
	{
		int mid = (l + r) / 2;
		build(id * 2, l, mid);
		build(id * 2 + 1, mid + 1, r);
		tree[id].sum = tree[id * 2].sum + tree[id * 2 + 1].sum;
	}
}
int query(int id, int l, int r)
{
	if (tree[id].left == l && tree[id].right == r)
		return tree[id].sum;
	int mid = (tree[id].left + tree[id].right);
	if (r <= mid)	return query(id * 2, l, r);
	else if (l > mid)	return query(id * 2 + 1, l, r);
	else	return query(id * 2, l, mid) + query(id * 2 + 1, mid + 1, r);
}
void update(int id, int l, int r, int delta)
{
	if (tree[id].left > r || tree[id].right < l)
		return;
	if (tree[id].left >= l && tree[id].right <= r)
	{
		tree[id].sum += (tree[id].right - tree[id].left + 1) * delta;
		tree[id].delta += delta;
		return;
	}
	if (tree[id].delta)
	{
		tree[id * 2].sum += (tree[id * 2].right - tree[id * 2].left + 1) * tree[id].delta; // 增量下传左子树
		tree[id * 2 + 1].sum += (tree[id * 2 + 1].right - tree[id * 2 + 1].left + 1) * tree[id].delta; // 增量下传右子树
		tree[id * 2].delta += tree[id].delta;
		tree[id * 2 + 1].delta += tree[id].delta;
		tree[id].delta = 0; 
	}
	update(2 * id, l, r, delta);
	update(2 * id + 1, l, r, delta);
	tree[id].sum = tree[2 * id].sum + tree[id * 2 + 1].sum;
}
int main()
{
	
	scanf("%d%d",&n,&t);
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]); 
	build(1, 1, n); 
	int op, l, r, k;
	while (t--)
	{
		scanf("%d", &op);
		if (op == 1)
		{
			scanf("%d%d%d", &l, &r, &k);
			update(1, l, r, k);
		}
		else if (op == 2)
		{
			scanf("%d", &k);
			update(1, 1, 1, k);
		}
		else if (op == 3)
		{
			scanf("%d", &k);
			update(1, 1, 1, -k);
		}
		else if (op == 4)
		{
			scanf("%d%d", &l, &r);
			printf("%d\n", query(1, l, r));
		}
		else if (op == 5)
		{
			printf("%d\n", query(1, 1, 1));
		 } 
	}
	
	return 0;
}